International Baccalaureate: Maths
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Infinite Surds portfolio
show that the relationship between n and L can be represented by L= an The data begins to increase by a smaller amount about each consecutive n, suggesting that the data may be approaching as asymptote. As these values get very large, they will probably not get much higher than the value of a10, because there already appears to be almost horizontal trend.
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SL IA Type 1: Infinite Summation Portfolio
2 To further clarify how these values were obtained a sample calculation is shown below: , a screenshot highlighting the formula used in Excel to complete the rest is shown in figure 1 below. Figure 1: Showing how Excel was used to calculate Sn If we now take these values and create a graph of n vs. Sn we obtain the result shown in Figure 2 From this graph it is clearly seen that the summation Sn converges on the value 2.
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Matrix power. The internal assessment will focused on observing patterns of matrix powers which will be the main key to find the general expression of matrix powers.
Explanation: When matrix P is powered by 3 it gives a result of , when matrix P is powered by 4 it gives a result of, and when matrix P is powered by 5 it gives a result of . The pattern shown is that results have a common factor of such as 4 shown in matrix, 8 shown in matrix, and then16 shown in matrix.
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Triangular and Stellar Numbers
value of 'y' by Un and the value of 'x' by n, thus: STELLAR NUMBERS NUMBER OF DOTS TO S6 STAGE S1 S2 S3 S4 S5 S6 1 13 37 73 121 181 Thus, using finite difference: The most obvious pattern is that the 1st row all numbers and odd and the second row all are even. Also all these numbers are some multiples of 12 + 1, for example: 12 also turns out to be the half of 6. 6 STELLAR NUMBER AT STAGE S7 1 + 1 (12) + 2 (12) + 3 (12) + 4(12) + 5(12)
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Discover how to solve inverse functions both graphically and algebraically, whilst investigating their relations, properties and patterns.
2. a) The linear function, f(x) = 2x + 4 is reflected. So the points on the graph, (0,4) become (4,0). b) (3x  1)/(x + 2) is reflected and becomes, (2x + 1)/(3  x). This results in a mirror image of the original function. c) f(x) = x� is reflected resulting in the inverse function, g(x) = �Vx. 3. Using the linear function, f(x) = 4x + 8, it is clear that my results in Q2 are indeed correct, as they are confirmed by the inverse function of the above linear function. It is flipped resulting in g(x) =.
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Mathematics portfolio on Infinite Surd
However, after that, anan+1 value has been had no huge change which means that difference is close to 0. Apply Here is a proved formula a= a2=1+a a2a1=0 Use quadratic equation a= = 1.618033989 or 0.6180339887 However, the root cannot be the negative number So, ? 1.618033989 Consider another infinite surd where the first term is Repeat the entire process above to find the exact value for this surd.
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The Straight Line
Examples 1. Find the yintercept for the following equation. * 2. Find the yintercept for the following straight line. Slope/Gradient The slope (gradient) ultimately determines the 'steepness' or incline of a line, the higher the slope, the steeper the incline will be. For example, a horizontal line has a slope equal to zero while a line with an angle of 45o has a slope equal to one. The sign (positive or negative) of the slope is very important, as it determines whether the line slopes uphill or downhill.
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Simple hill cypher  using matrices to create a coded message.
Normally when you encrypt something, you would most commonly have "units" substituting your letter or sentences. The units could be in one's, two's or even three's. There are various ways of substitution cyphering, one way (the most common way), would be to use ROT13, an alphabet rotated over 13 steps. Simple Hill Cypher, is a method to solve and encrypt a message. The method consists of using substitution. Eg: A B C D E F G H I J K L M N O N O 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 14 15 P Q R S T U V W X Y Z .
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Ib math SL portfolio parabola investigation
A graph of the displayed functions is below this (Figure 2.2). An explanation for each significant situation is provided below Figure 2.2. Figure 2.1 Situation # 2 1 3 4 4 2 5 6 7 8 9 10 [tt2] Situation 3: This situation caused me to hypothesize the conjecture that, where A is the A value from the equation. In addition, g(x) was tangent to f(x) at the point (10,10). This meant that x2 and x3 were the same (10), and showed that the conjecture held true for tangents. Situation 6: A concave down parabola in the 1st quadrant still holds the conjecture, provided that the conjecture is changed to Situation 7: Irrational A values work for the conjecture.
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Logarithm Bases Math IA
2) Cleary, the pattern is noticeable as the sequence goes on. Therefore, I was able to find the term for each sequence, writing it in the form where . For the general logarithmic sequence, the term is the following. ..., That being the case, the nth term for the three examples of logarithmic sequences are the following: 1) 2) 3) To justify my answer using technology, I used excel to verify that the pattern continues (see appendix 1). By carrying out the pattern to where n=50, I was able to confirm that the pattern continues.
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Portfolio: Body Mass Index
To be more sure, I drew a simple Gaussian function over the graph, which I already have had. The new graph looks as follows (the red line is the function that the given data form and the black line is the Gaussian function): The exact equation of the Gaussian function, which fits the data graph, is: y =, where A=6,933 B= 5,510 C=8.846 D=22.15 It must be noted that it is highly possible the Gaussian function to be inappropriate if the data is for elder people.
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Investigating graph of trigonometric function
If however we reduce the value, then the curve will vertically contract with an amplitude of the same value. Finally, if we inverse the sign of the amplitude for example we change 'a' into 'a', then the curve will reflect through the centreline. If we consider an infinitely extended graph, then we could say that the values of 'a' could be infinite apart from zero as the curve would vertically stretch or contract at any number apart from zero.
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infinite surds
One day Jimmy walked Kandee home from school and Kandee's dad saw them holding hands. He went crazy and ran after Jimmy. Luckily Jimmy was too fast and Kandee's dad didn't catch him. Kandee explained that the teacher told her to walk the poor blind penguin home. The funny thing is her dad actually believed her even though it looked like he knew what direction to run in. From then on Jimmy had to act blind in order to come to their house everyday.
 Word count: 800

Crows Dropping Nuts
The following table shows the number of average drops it takes at a certain height to break the nut. Height of drop(m) 1.7 2.0 2.9 4.1 5.6 6.3 7.0 8.0 10.0 13.9 Number of drops 420 21.0 10.3 6.8 5.1 4.8 4.4 4.1 3.7 3.2 This is a graphical representation of the large nuts data using TI InterActive. Relationship of Height of the drop vs. Number of drops for the Large Nuts The variables in this data are that the height of the nut dropped affected the frequency, and this variable is put into an average.
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IB Design Tech Design Cycle
For the bridge, I thought that tying 3 straws to form a bundle and using 4 bundles to form a square and 2 other bundles to cross inside the square will create a strong base. Every 5 inches I will place a base so that the building will hold up straight. My goal for the tower was to be 2 feet tall or even taller. For the joints there should be a 40 degrees triangle formed by 2 bundles cutting across the side of the building.
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Volumes of Cones
Therefore, using the Pythagorean Theorem, we can find h. r2 = h2 + R2 r2 = h2 + (r?/2?)2 h2 = r2  (r?/2?)2 h = ? [r2  (r2?2/4?2)] Therefore, substituting the values for rbase and h, we can find the volume of the cone. V = 1/3 x height x base area V = 1/3 ? ? [r2  (r2?2/4?2)] ? ? (r2?2/4?2) 2. By using the substitution x = ?/2? , express the volume as a function of x. x = ?/2? ? = x2? V = 1/3 ? ? [r2  (r2?2/4?2)] ? ? (r2?2/4?2) V = 1/3 ? ? [r2  (r2(x2?)2/4?2)] ? ?
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MATRIX BINOMIALS. In this investigation, we will identify a general statement by examining the patterns of the matrices.
Thus, with these patterns, the following expression can be suggested: The matrices X and Y can now be used to form two new matrices A and B. Here, we will use a and b as constants for the matrices A and B, respectively. And hence the following: , Now, the different values of a and b can be used to calculate the values of And therefore, , , , , With the patterns from these matrices, we can determine the expressions for matrices A and B by considering its integer powers: We will now investigate a new matrix, Here, we
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IB Math HL Portfolio Type 1 (Ratio)
Area B: y=xn dx = = = = units2 Therefore, the ratio of the areas A and B for the function y=xn is: A : B = : = n : 1 Conjecture for the ratio of the areas A and B for the function y=xn is A : B = n : 1. Now, try and test the conjecture for other subsets of the real numbers. When n=3, Area A: y=x3 x= dy =[ ] =[] =( = units2 Area B: y=x3 dx = = = = units2 Therefore, the ratio of the areas A and B in the
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Shady Areas
+ 1/2(a x d) = area of trapezium bc + 1/2(e  b)c c [ b + 1/2(e  b) ] c [ 2b/2 + (eb)/2 ] = c[ (e + b)/2 ] x = g(x) for x� + 3= g(x) 0.2 3.04 0.4 3.16 0.6 3.36 0.8 3.64 1.0 4 0.0 3 Graph 1 area = 0.5 [ (3.25 + 3)/2 ] = 1.5625 = area of first trapezium 0.5 [ (4 + 3.25)/2 ] = 1.8125 = area of second trapezium 1.5625 + 1.8125 = 3.375 For Graph 1 the sum of the area of the two trapeziums gives a result of 3.375 By increasing the number of trapeziums we can gain a more accurate estimate of the area under the curve.
 Word count: 880

Portfolio SL  Matrix
* The entries double for every higher power of Y, i.e.: Y2= 2Y Y3= 4Y= 22Y Y4= 8Y = 23Y follows Yn=2n1Y Again I will test is with random numbers using GDC: Y8= Y8=281= Y16= Y16=2161 Yn=2n1Y formula valid for all natural numbers N; = {1,2,3,...} as it is shown in examples. * In the case of (X+Y)= 2I (X+Y)= = =2I (X+Y)n= n = n = = 2n In this case we have two slightly different formulas, the first one includes the sum of (X+Y)
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IB Math Portfolio Investigating Ratios
Let , Finding the ratio of area A: area B: The ratio of area A: area B is 4:1. Let , Finding the ratio of area A: area B: The ratio of area A: area B is 10:1. From the findings above, it is evident that when , the ratio of area A: area B is always n : 1 when n is a positive integer more than 0. To further investigate my conjecture, test the conjecture for other subsets of the real numbers: Fractions: Let , Finding the ratio of area A: area B: The ratio of area A: area B is :1, which means that the conjecture is true to fraction values of n.
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Matrix Binomials Portfolio
B=bX > b Let: a = 3 and b = 4 A2 = A3 = A4 = B2 = B3 = B4 = Therefore: (A+B) = (A+B)2 = (A+B)3 = (A+B)4 = Expressions for An, Bn and (A+B)n An = Bn = (A+B)n = n > 0, let: W = any 22 matrix, Wn = > It is not possible to divide an integer by a matrix, n < 0 does not exist n=0 For any matrix where
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Infinite Surds Coursework
However this increase is not proportional to the increase of n, an seems to be increasing towards 1.62. Once n reaches 28 an ceases to increase, remaining stable at 1.618033988749890. This suggests that as n becomes very large an  an+1 = 0 As such, we can conclude that the exact value for this infinite surd is 1.618033988749890. Consider another infinite surd: Find the formula for an+1 in terms of a a1 = a2 = a2 = a3 = a3 = an+1 = an = Calculate the decimal values of the first ten terms of the sequence a1 = 1.847759065
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Infinite surds Maths Portifolio
This can be understood by: an  an+1 as n gets very large. lim(an  an+1) ? 0 When n gets very large and approaches infinity, the value approaches 0 because the difference between these two values become very small. This is a way to find the exact value of this infinite surd: an+1 can also be written as an so therefore an = an2 = 1+an an2  an  1 = 0 abc formula: a = a1 = 1.618 a2 = 0.618 Since the value has to be a positive number, the exact value of this infinite surd is 1.618.
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